Answer:
The length of the wire used is [tex]L = 12.85\cdot d[/tex], where [tex]d[/tex] is the diameter of the semicircle.
Step-by-step explanation:
Let be a wire of length [tex]L[/tex]. Geometrically speaking, the perimeter of a semicircle is:
[tex]s = \left(\frac{\pi}{2}+1 \right)\cdot D[/tex] (1)
Where:
[tex]s[/tex] - Perimeter.
[tex]D[/tex] - Diameter.
Given that wire was used to create a two large semicircles and a small semicircle and that diameter of the large semicircle is twice the diameter of the small semicircle, we have the following formula:
[tex]L = 2\cdot \left(\frac{\pi}{2}+1\right)\cdot (2\cdot d) + \left(\frac{\pi}{2}+1 \right)\cdot d[/tex]
[tex]L = 5\cdot \left(\frac{\pi}{2}+1 \right)\cdot d[/tex]
[tex]L = 12.85\cdot d[/tex]
Where [tex]d[/tex] is the diameter of the semicircle.
